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Very little interest in the original version of this question so I've rejigged it hoping for a more positive response.

I'm trying to use the geodesic deviation equation$$\frac{D^{2}\xi^{\mu}}{D\lambda^{2}}+R_{\phantom{\mu}\beta\alpha\gamma}^{\mu}\xi^{\alpha}\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}=0$$ to show that on the surface of a unit sphere two particles separated by initial distance $d$ , starting from the equator and travelling north (ie on lines of constant $\phi$) will have a separation $s$ given by$$s=d\sin\theta.$$ This is similar to Geodesic devation on a two sphere except that question was solved using simple spherical geometry.

My plan is find $\frac{D^{2}\xi^{\mu}}{D\lambda^{2}}$ first by using the absolute derivative $$\frac{DV^{\alpha}}{d\lambda}=\frac{dV^{\alpha}}{d\lambda}+V^{\gamma}\Gamma_{\gamma\beta}^{\alpha}\frac{dx^{\beta}}{d\lambda}.$$ Then take the second derivative of this. Next find $\frac{D^{2}\xi^{\mu}}{D\lambda^{2}}$ by calculating the Riemann tensor part$$R_{\phantom{\mu}\beta\alpha\gamma}^{\mu}\xi^{\alpha}\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}.$$ And then try to juggle the results to show the separation $s=\xi^{\phi}$ as a function of $\theta$.

The line element for spherical coordinates $$l^{2}=dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2}$$ for a great circle of constant $\phi$ on a sphere of unit radius reduces to $$dl^{2}=d\theta^{2}$$ giving $\frac{d\theta}{dl}=\frac{d\theta}{d\lambda}=1$ and $\frac{d\phi}{dl}=\frac{d\phi}{d\lambda}=0$.

The absolute derivative for $\Gamma_{\theta\phi}^{\phi}=\Gamma_{\phi\theta}^{\phi}=\frac{\cos\theta}{\mathbf{\mathbf{\sin\theta}}}$ is$$\frac{D\xi^{\phi}}{d\lambda}=\frac{d\xi^{\phi}}{d\lambda}+\xi^{\theta}\Gamma_{\theta\phi}^{\phi}\frac{d\phi}{d\lambda}+\xi^{\phi}\Gamma_{\phi\phi}^{\phi}\frac{d\phi}{d\lambda}+\xi^{\theta}\Gamma_{\theta\theta}^{\phi}\frac{d\theta}{d\lambda}+\xi^{\phi}\Gamma_{\phi\theta}^{\phi}\frac{d\theta}{d\lambda}=\frac{d\xi^{\phi}}{d\lambda}+\xi^{\phi}\frac{\cos\theta}{\mathbf{\mathbf{\sin\theta}}}.$$

And for $\Gamma_{\phi\phi}^{\theta}=\sin\theta\cos\theta$ is $$\frac{D\xi^{\theta}}{d\lambda}=\frac{d\xi^{\theta}}{d\lambda}+\xi^{\phi}\Gamma_{\phi\phi}^{\theta}\frac{d\phi}{d\lambda}+\xi^{\phi}\Gamma_{\phi\theta}^{\theta}\frac{d\theta}{d\lambda}+\xi^{\theta}\Gamma_{\theta\phi}^{\theta}\frac{d\phi}{d\lambda}+\xi^{\theta}\Gamma_{\theta\theta}^{\theta}\frac{d\theta}{d\lambda}=\frac{d\xi^{\theta}}{d\lambda}.$$

However, $$\frac{D\xi^{\phi}}{d\lambda}=\frac{d\xi^{\phi}}{d\lambda}+\xi^{\phi}\frac{\cos\theta}{\mathbf{\mathbf{\sin\theta}}}$$ doesn't look right as it blows up when $\theta=0$. Any suggestions where I might be going wrong?

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    $\begingroup$ There's some implicit summations going on (Einstein summation convention): $V^{\gamma}\Gamma_{\gamma\beta}^{\alpha}\frac{dx^{\beta}}{d\lambda}$ means $\sum_{\gamma,\beta}V^{\gamma}\Gamma_{\gamma\beta}^{\alpha}\frac{dx^{\beta}}{d \lambda}$, with the sum taken over repeated indices. $\endgroup$ – Holographer Sep 3 '14 at 8:53
  • $\begingroup$ Whoops. I've now amended the question to show my results after summing over the indices. $\endgroup$ – Peter4075 Sep 3 '14 at 10:13
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I've since found out that where I'm going wrong is that I don't need to find the absolute derivative. I've been told on another physics forum that this problem is framed in terms of Riemann normal coordinates that makes it OK to assume $\frac{D^{2}\xi^{\mu}}{dt{}^{2}}=\frac{d^{2}\xi^{\mu}}{dt{}^{2}}$. Apparently, this is because the distance the cars travel along their separate geodesics on the sphere is a linear function of time $t$.

Anyway, the calculation is as follows.

The distance each car travels north from the equator along its $\phi=$ constant geodesic is $vt$. The angle car-centre of sphere-equator is $vt/r=vt$. Spherical coordinate $\theta=\left(\pi/2\right)-vt$.

The equation of geodesic deviation is $$\frac{d^{2}\xi^{\mu}}{dt{}^{2}}+R_{\phantom{\mu}\beta\alpha\gamma}^{\mu}\xi^{\alpha}\frac{dx^{\beta}}{dt}\frac{dx^{\gamma}}{dt}=0.$$

In effect we want to find $s=\xi^{\phi}$ as a function of $t$.

The non-zero Riemann components for a unit sphere are: $R_{\phantom{\theta}\phi\theta\phi}^{\theta}=\sin^{2}\theta$, $R_{\phantom{\theta}\phi\phi\theta}^{\theta}=-\sin^{2}\theta$, $R_{\phantom{\theta}\theta\theta\phi}^{\phi}=-1$, $R_{\phantom{\theta}\theta\phi\theta}^{\phi}=1$.

Let $u^{\sigma}\equiv\frac{dx^{\sigma}}{dt}$. Then expand out Riemann components for a unit sphere to get:

$$\frac{d^{2}\xi^{\theta}}{dt{}^{2}}=\left(\sin^{2}\theta\right)\left(u^{\phi}u^{\theta}\right)\xi^{\phi}-\left(\sin^{2}\theta\right)\left(u^{\phi}u^{\phi}\right)\xi^{\theta}$$.But as the cars are moving along lines of constant $\phi$, $$u^{\phi}=\frac{d\phi}{dt}=0$$ and we get$$\frac{d^{2}\xi^{\theta}}{dt{}^{2}}=0$$.

Next set $\mu=\phi$, expand out the Riemann components to give

$$\frac{d^{2}\xi^{\phi}}{dt{}^{2}}=\xi^{\theta}\left(u^{\theta}u^{\phi}\right)-\xi^{\phi}\left(u^{\theta}u^{\theta}\right).$$

As $\theta=\left(\pi/2\right)-vt$, $u^{\theta}=\frac{d\theta}{dt}=-v$ and we get$$\frac{d^{2}\xi^{\phi}}{dt{}^{2}}=-v^{2}\xi^{\phi}.$$

Using the wizardry of Wolfram Alpha, the solution to $$\frac{d^{2}y}{dx^{2}}=-ky$$ is$$y=A\sin\left(x\sqrt{k}\right)+B\cos\left(x\sqrt{k}\right).$$ Substituting $v^{2}=k$, $t=x$ and $\xi^{\phi}=y$ gives$$\xi^{\phi}=A\sin\left(vt\right)+B\cos\left(vt\right).$$

To find the constants $A$ and $B$, we note that when the distance travelled $vt=0$ $\xi^{\phi}=d$, giving$$\xi^{\phi}=d=A\sin\left(0\right)+B\cos\left(0\right)$$ giving$$B=d,$$ and therefore$$\xi^{\phi}=d\cos\left(vt\right).$$

This is the same answer as calculated in Geodesic devation on a two sphere. In spherical coordinates this is equivalent to $$\xi^{\phi}=d\sin\left(\theta\right).$$

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